Find whether $\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $ is geometric and convergent and if so, state the value of the sum

Question

$$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $$

First I tried to find the ratio by doing $\frac{a_{n+1}}{a_n}$. Many calculations were made, and the result was $24/25$, so it is geometric. This is between -1 and 1, and therefore the series is convergent.

Then I did $$\sum_{d=1}^\infty \frac{a_1}{1-24/25} = (...) = 675$$

Is this correct?

Answer 1

$$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} =\frac{9.25}{8}\sum_{n=1}^{\infty } \frac{8^n.3^n}{25^n}=\frac{225}{8}\frac{\frac{24}{25}}{1-\frac{24}{25}}=675$$

So your answer is true.

Answer 2

HINT

You have $$ \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} = \frac{2^{-3} 3^2}{5^{-2}} \frac{2^{3n} 3^n}{5^{2n}} = \frac{25 \cdot 9}{8} \frac{8^n 3^n}{25^n} = \frac{25 \cdot 9}{8} \left(\frac{24}{25}\right)^n $$